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$(x^*Ay)^2 \leq (x^*Ax)(y^*Ay)$

Does this inequality hold for positive semi-definite matrix A? I wonder if this is equivalent to Cauchy-Schwarz Inequality. I tried to diagonalize A but it still half-way towards Cauchy-Schwarz Inequality.

Any thoughts on this inequality? Am I going on the right direction?

Thanks a lot!

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Complex or real? If complex, you need $(x^*Ay)(y^*Ax)$ on the left. –  Jonas Meyer Aug 27 '11 at 16:39
    
@Jonas Meyer Real –  BVFanZ Aug 27 '11 at 17:00

3 Answers 3

up vote 4 down vote accepted

The Cauchy-Schwarz inequality applies to any semi-definite inner product, including the one on $\mathbb{R}^n$ (as column vectors) defined by $\langle x,y\rangle=x^{\mathrm T}Ay$.

One of the simplest proofs in the real case is to consider $p(t)=\langle x-ty,x-ty\rangle=\langle x,x\rangle -2\langle x,y\rangle t +\langle y,y\rangle t^2$. Then $p(t)\geq 0$ for all $t\in \mathbb R$, and $p$ is a quadratic polynomial in $t$, so its discriminant must be nonpositive. Rearranging this discriminant inequality yields your inequality.

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The matrix $A$ has a Cholesky decomposition: $A=L^* L$. Thus

$$x^* A y = (Lx)^* Ly = (Lx) \cdot (Ly),$$ where by $\cdot$ we mean the usual complex inner product: $a \cdot b = \sum_i \overline{a_i} b_i$. Now applying Cauchy-Schwarz,

$$ |x^* A y|^2 \leq ||Lx||_2^2 ||Ly||_2^2 = (x^* L^* L x) (y^* L^* L y) = (x^* A x) (y^* A y)$$

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1  
BVFanZ has clarified in a comment that only the real case is being considered. In the complex case, you should have $|x^*Ay|^2$ rather than $(x^*Ay)^2$. –  Jonas Meyer Aug 27 '11 at 17:03
    
@Jonas Meyer - indeed, thanks. –  alexx Aug 27 '11 at 17:05

Yes, it is equivalent to CS inequality. W.l.o.g, asumme $A$ is a diagonal matrix with positive diagonal entries, $A=diag(a_1,\cdots, a_n)$. you will see that $\left(\sum \sqrt{a_i}x_i\sqrt{a_i}y_i\right)\le \left(\sum (\sqrt{a_i}x_i)^2\right)\left(\sum (\sqrt{a_i}y_i)^2\right) $ is in the same form of CS inequality.

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